Question

use for questions 10 - 12: stephanie recorded the number of miles that she ran during various days in september.

1. what is the rate of change in miles ran from september 6th to september 15th?
2. what is the rate of change in miles ran from september 15th to september 18th?
3. what is the rate of change in miles ran from september 18th to september 21st?

Explanation:

To solve these rate - of - change problems, we use the formula for the rate of change (slope) between two points \((x_1,y_1)\) and \((x_2,y_2)\), which is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here, \(x\) represents the number of days and \(y\) represents the number of miles run.

Question 10: Rate of change from September 6th to September 15th

Step 1: Identify the points

On September 6th (\(x_1 = 6\)), the number of miles run \(y_1=4\). On September 15th (\(x_2 = 15\)), the number of miles run \(y_2 = 6\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{6 - 4}{15 - 6}=\frac{2}{9}\approx0.22\) miles per day.

Question 11: Rate of change from September 15th to September 18th

Step 1: Identify the points

On September 15th (\(x_1 = 15\)), \(y_1 = 6\). On September 18th (\(x_2=18\)), \(y_2 = 5\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{5 - 6}{18 - 15}=\frac{- 1}{3}\approx - 0.33\) miles per day.

Question 12: Rate of change from September 18th to September 21st

Step 1: Identify the points

On September 18th (\(x_1 = 18\)), \(y_1=5\). On September 21st (\(x_2 = 21\)), \(y_2=9\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{9 - 5}{21 - 18}=\frac{4}{3}\approx1.33\) miles per day.

Question 10 Answer: \(\frac{2}{9}\) (or approximately \(0.22\)) miles per day

Question 11 Answer: \(-\frac{1}{3}\) (or approximately \(- 0.33\)) miles per day

Question 12 Answer: \(\frac{4}{3}\) (or approximately \(1.33\)) miles per day

Answer:

To solve these rate - of - change problems, we use the formula for the rate of change (slope) between two points \((x_1,y_1)\) and \((x_2,y_2)\), which is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here, \(x\) represents the number of days and \(y\) represents the number of miles run.

Question 10: Rate of change from September 6th to September 15th

Step 1: Identify the points

On September 6th (\(x_1 = 6\)), the number of miles run \(y_1=4\). On September 15th (\(x_2 = 15\)), the number of miles run \(y_2 = 6\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{6 - 4}{15 - 6}=\frac{2}{9}\approx0.22\) miles per day.

Question 11: Rate of change from September 15th to September 18th

Step 1: Identify the points

On September 15th (\(x_1 = 15\)), \(y_1 = 6\). On September 18th (\(x_2=18\)), \(y_2 = 5\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{5 - 6}{18 - 15}=\frac{- 1}{3}\approx - 0.33\) miles per day.

Question 12: Rate of change from September 18th to September 21st

Step 1: Identify the points

On September 18th (\(x_1 = 18\)), \(y_1=5\). On September 21st (\(x_2 = 21\)), \(y_2=9\).

Step 2: Apply the rate - of - change formula

Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we substitute the values:
\(m=\frac{9 - 5}{21 - 18}=\frac{4}{3}\approx1.33\) miles per day.

Question 10 Answer: \(\frac{2}{9}\) (or approximately \(0.22\)) miles per day

Question 11 Answer: \(-\frac{1}{3}\) (or approximately \(- 0.33\)) miles per day

Question 12 Answer: \(\frac{4}{3}\) (or approximately \(1.33\)) miles per day